Question

answer parts (a) through (c)
(a) is the statement -|x| ≤ x sin(\frac{2}{x}) ≤ |x| true for all x ≠ 0? choose the correct answer below.
no
yes

Explanation:

Step1: Recall the range of sine function

The range of $\sin t$ for any real - valued $t$ is $[- 1,1]$. Let $t=\frac{2}{x}$ ($x
eq0$). Then $-1\leqslant\sin(\frac{2}{x})\leqslant1$.

Step2: Multiply by $|x|$

Since $|x|>0$ for $x
eq0$, when we multiply the inequality $-1\leqslant\sin(\frac{2}{x})\leqslant1$ by $|x|$, we get $-|x|\leqslant|x|\sin(\frac{2}{x})\leqslant|x|$. Also, note that $x\sin(\frac{2}{x})$ and $|x|\sin(\frac{2}{x})$ have the same absolute - value because $|x\sin(\frac{2}{x})| = |x|\cdot|\sin(\frac{2}{x})|$ and the sign of $x$ is accounted for in the absolute - value operation. So, $-|x|\leqslant x\sin(\frac{2}{x})\leqslant|x|$.

Answer:

Yes